Another approach: If there is any one out of 3 variables which is divisble by 2, then it must be equal to 2. If a=2 or b=2 we easily infer that the equation doesn't have any prime solutions! If c=2 then we will have the inequality: a^4<=16 then easily to infer that a=2. But then b=0(contradicts to th assumption) Now, if there is not any variables which are divisible by 2, then a^4+b^4 module 4 is equal to 2. But c^4 module 4 is equal to 1(contradiction since a^4+b^4=c^4) From 2 cases above, we are done!

Is it valid in a contest? Absolutely

Quote: Fermat's last Theorem, since all primes lie in the Natural Numbers You are absolutely right but it was for 2.nd grades Maybe just a few students know about Fermat's last theorem.

CatalystOfNostalgia wrote: Thus $ p^{2} = x^{2} - y^{2},q^{2} = 2xy$. The second equation gives $ q = 2,xy = 2$, so $ x = 2,y = 1$. I'm not sure how you concluded this. We can have $ x = 2a^2, y = b^2$ where $ \gcd(a, b) = 1$ and $ b$ is odd. The proof that this equation has no solution is usually done by infinite descent.

t0rajir0u wrote: CatalystOfNostalgia wrote: Thus $ p^{2} = x^{2} - y^{2},q^{2} = 2xy$. The second equation gives $ q = 2,xy = 2$, so $ x = 2,y = 1$. I'm not sure how you concluded this. We can have $ x = 2a^2, y = b^2$ where $ \gcd(a, b) = 1$ and $ b$ is odd. The proof that this equation has no solution is usually done by infinite descent. You're right, t0raijru! .But it isn't essential in this case since the restriction:"prime numbers" make the problem more easier than the original problem. That maybe is suitable for the level of student in that age.

I'd assume Fermat's Last Theorem wouldn't be accepted as it trivializes the problem.

Quote: I'd assume Fermat's Last Theorem wouldn't be accepted as it trivializes the problem. It could be interesting if someone solved problem by Fermat's Last theorem but I assume that nobody of contestants know about it ( 2.nd grade) Most probabaly solution based on this theorem would be accepted!

$ p^4=(r^2+q^2)(r+q)(r-q)$. Note that $ r^2+q^2>r+q>r-q$. Then $ r^2+q^2=p^3$, $ r+q=p$, $ r-q=1$. Then $ r=\frac{p+1}{2}$, $ q=\frac{p-1}{2}$, in the first equation, $ (p-1)(2p^2+p+1)=0$ which has no prime solution.

There is a solution in Aops volume 2:

A very easy approach of this problem: If p and q are odd, then r is even, but if r is even, since r is prime, r=2.Here, it's easy to see that there aren't solutions for p^4+q^4. So, at least one of this primes is even.Wlog, let p this prime, so p=2.Now, we have:16=(r²2+q²)(r²-q²) . Here, it's again easy to see that there aren't solutions.So we are done. Bye bye

Maybe I'm being stupid, but just Fermats last theorem????

The point is to prove this result independently. The case $ n = 4$ was known even to Fermat, and the case for $ x, y, z$ prime is even easier.

1. Ok, fine. 2. This is a 2ND GRADE competition?? I can't really see how 2nd graders would be doing this.

Perhaps in Bosnia, Grade 2 does not equate to Grade 2 where you're from. In California, Grade 2 is generally ages 7-8. This does seem a bit difficult for that age bracket, but again, grade 2 in Bosnia is most likely seen differently.

Grade 2 (more precisely, high school 2nd grade) in Bosnia is equivalent to grade 10, i.e. 15-16 year old students.

That would explain

1)if $p,q>2=>r=2$ no solutions)) 2)if either $p$ or $q$ equals $2=>16=(r-q)(r+q)(r^2+q^2)$ no solutions)) very easy one)

t0rajir0u wrote: CatalystOfNostalgia wrote: Thus $ p^{2} = x^{2} - y^{2},q^{2} = 2xy$. The second equation gives $ q = 2,xy = 2$, so $ x = 2,y = 1$. I'm not sure how you concluded this. We can have $ x = 2a^2, y = b^2$ where $ \gcd(a, b) = 1$ and $ b$ is odd. The proof that this equation has no solution is usually done by infinite descent. Why do you think Pythagorean triples are wrong since one can take $a^2=p^4,b^2=q^4,c^2=r^4$ and get Pythagoras equation ????? That method is perfectly ok!

$p, q, r\in\mathbb{P}$ $p^2=u^2–v^2$ $q^2=2xy$ not prime since $\sqrt{2}.k$ $r^2=u^2+v^2$ clearly not primes....

Work in modulo 6 and the answer will be revealed.

delegat wrote: Prove that equation $ p^{4}+q^{4}=r^{4}$ does not have solution in set of prime numbers. We can prove by "descente infinite" that the equation $ p^{4}+q^{4}=r^{2}$ have not positive integer solution, so the equation $ p^{4}+q^{4}=r^{4}$ have not too positive integer solutions, even more prime solutions!

Yea somewhat like @rikabcd123456 says can't we simply work modulo $4$ ? $(p,q,r) \equiv \pm 1 \pmod 4 \implies p^4,q^4,r^4 \equiv 1 \pmod 4$.Then the LHS $\equiv 2 \pmod 4$ whereas RHS $\equiv 1 \pmod 4$.We are done. @targo all primes are 4k+1 or 4k+3 except 2.

AYG wrote: Yea somewhat like @rikabcd123456 says can't we simply work modulo $4$ ? $(p,q,r) \equiv \pm 1 \pmod 4 \implies p^4,q^4,r^4 \equiv 1 \pmod 4$.Then the LHS $\equiv 2 \pmod 4$ whereas RHS $\equiv 1 \pmod 4$.We are done. How you got $p,q,r\equiv \pm 1\pmod{4}$ ? You need to consider that , as , $p,q<r$ , we will have one of $p$ or, $q$ equals to $2$,to maintain the parity.So, WLOG if you take $p<q$ then,$p=2$ .

Is this ok? We can clearly see that $p=q=2$ is not a solution. $q^4=(r^2+p^2)(r+p)(r-p)$ So if $p\neq 2 $ we get that no part of right side is equal to $1$ and no part is negative. Thus we have that the right side is "made" of $q^2, q, q$. So one part in brackets is equal to one of this. But that would mean that 2 of them would be equal which makes no sense because we have prime numbers which are greater than $1$. So that doesn't work. Now if we have $p=2$ we just switch $p$ and $q$ and we get the same equation.

delegat wrote: Prove that equation $ p^{4}+q^{4}=r^{4}$ does not have solution in set of prime numbers. Lemma: Let $\left(a,b,c\right)$ be a Pythagorean triple. Then $abc$ is a multiple of $60$. Proof: We begin by proving that one of $a$,$b$,$c$ is a multiple of $3$. Assume the contrary, then $c^2=a^2+b^2=2\left(\operatorname{mod}3\right)$. This is impossible hence one of $a$,$b$,$c$ is a multiple of $3$. Now take the numbers modulo $8$. If no $a$,$b$,$c$, then $c^2=1,4\left(\operatorname{mod}8\right)$ and $a^2+b^2=2,5,0\left(\operatorname{mod}8\right)$. Impossible, hence one of $a^2$, $b^2$, $c^2$ is a multiple of $8$ which implies one of $a$,$b$,$c$ is a multiple of $4$. Finally, take the numbers modulo $5$. Assume none are a multiple of $5$. Then $c^2=1,4\left(\operatorname{mod}5\right)$ and $a^2+b^2=2,3,0\left(\operatorname{mod}5\right)$. Hence one of $a$,$b$,$c$ is a multiple of $5$. $60=3\cdot4\cdot5$ hence $abc$ is a multiple of $60$. From the lemma, $p^2q^2r^2$ is a multiple of $60$ and hence $pqr$ is a multiple of $30$. As they are primes, the only possible solution would be $\left(p,q,r\right)=\left(2,3,5\right)$. These do not work hence there are no solutions.

If $p, q, r$ are all odd primes, then there are no solutions by simply taking both sides mod 4 (LHS is $2 \pmod 4$ while RHS is $1 \pmod 4$). Hence at least one of $p, q, r$ is $2$. Clearly, it is not possible for $p=q=r=2$. It is also easy to check that we cannot have $2$ of the 3 variables be equal to 2 and the last one be an odd prime. This means that exactly one of the 3 variables must be equal to 2. If $r=2$, this is not possible as we get $p^4+q^4=16$, and the minimal value of the left side is $3^4+3^4=162 > 16$. So, WLOG let $p=2$. Therefore, $$16+q^4=r^4 \implies 16=r^4-q^4 \implies 16=(r^2+q^2)(r+q)(r-q).$$However, the minimal value of $r^2+q^2$ is $3^2+3^2=18$, so there are no solutions here. Since we've covered all the possibilities, we can conclude that there are no solutions.

Fermat's last theorem and directly it has no solution

Let $p, q, r$ are all $>3$ then must $p^4+q^4\equiv 2\pmod 6$ but $r^4\equiv 1\pmod 6$ done. Rest is case check of $\{2,3\}$.

Groupsolved with IAmTheHazard, sixoneeight, CT17, CT18, megarnie, OronSH, MathLuis, pikapika007, Martin2001, naman12, IAteTheHazard, tienxion, Evil Chin Inc Follows immediately by Fermat's Last Theorem.

P= r^2 - q^2 and P^3 = r^2 +q^2

I think this was relatively simple - but maybe I did something wrong?

It follows immediately by Fermat's Last Theorem. Or any 4th power of a prime is always 1 mod 3 besides if the prime is 3. Hence contradiction so one of the primes is 3. If r=3 no solutions. quick check yeilds no 4th powers have difference of 81.

How do you instantly notice that any prime number to 4th power is always $1 \pmod 3$ expect for 3?

PaperMath wrote: How do you instantly notice that any prime number to 4th power is always $1 \pmod 3$ expect for 3? any number to a 4th power mod 3 thats not 0 is equal to 1, since any number squared mod 3 = 1.

qwerty123456asdfgzxcvb wrote: PaperMath wrote: How do you instantly notice that any prime number to 4th power is always $1 \pmod 3$ expect for 3? any number to a 4th power mod 3 thats not 0 is equal to 1, since any number squared mod 3 = 1. Uhhhhh that doesn’t answer my question. That is literally what im asking, how people can identify those relationships better. Can someone also check my solution? -

The mod $3$ argument is nearly identical to what you did for mod $5$. In general, if you reduce $a^k$ mod $p$, you'll get $1$ if $p-1\mid k$. So a good place to start is with primes $p$ for which $p-1$ is a factor of the exponent.

hsiljak wrote: Grade 2 (more precisely, high school 2nd grade) in Bosnia is equivalent to grade 10, i.e. 15-16 year old students. I mean there are a significant amount of 10 graders that know fermat's last (like me)

Awesome_Twin1 wrote:

Don't forget to account for the cases of $r=2,3$ as these are different!!